Problem 24 - Chemistry: The Central Science in SI Units, Global Edition

Question

For each of the following gas-phase reactions, write the rate expression in terms of the appearance of each product and disappearance of each reactant: (a) \(\mathrm{O}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{O}_{2}(g)+\mathrm{H}_{2}(g)\) (b) \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) (c) \(2 \mathrm{C}_{2} \mathrm{H}_{2}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{2}(g) \longrightarrow \mathrm{C}_{3} \mathrm{H}_{6}(g)+\mathrm{NH}_{3}(g)\)

Step-by-Step Solution

Answer

The rate expressions for the given reactions are: (a) Rate = -\(\frac{d[O_3]}{dt}\) = -\(\frac{d[H_2O]}{dt}\) = \(\frac{1}{2}\) \(\frac{d[O_2]}{dt}\) = \(\frac{d[H_2]}{dt}\) (b) Rate = -\(\frac{1}{4}\) \(\frac{d[NH_3]}{dt}\) = -\(\frac{1}{5}\) \(\frac{d[O_2]}{dt}\) = \(\frac{1}{4}\) \(\frac{d[NO]}{dt}\) = \(\frac{1}{6}\) \(\frac{d[H_2O]}{dt}\) (c) Rate = -\(\frac{1}{2}\) \(\frac{d[C_2H_2]}{dt}\) = -\(\frac{1}{5}\) \(\frac{d[O_2]}{dt}\) = \(\frac{1}{4}\) \(\frac{d[CO_2]}{dt}\) = \(\frac{1}{2}\) \(\frac{d[H_2O]}{dt}\) (d) Rate = -\(\frac{d[C_3H_7NH_2]}{dt}\) = \(\frac{d[C_3H_6]}{dt}\) = \(\frac{d[NH_3]}{dt}\)

Step 1: Analyzing Coefficients

In this reaction, all reactants and products have a coefficient of 1, except for O2 which has a coefficient of 2.

Step 2: Rates of Appearance and Disappearance

Using the coefficients, the rate expression can be written as: Rate = -\(\frac{1}{1}\) \(\frac{d[O_3]}{dt}\) = -\(\frac{1}{1}\) \(\frac{d[H_2O]}{dt}\) = \(\frac{1}{2}\) \(\frac{d[O_2]}{dt}\) = \(\frac{1}{1}\) \(\frac{d[H_2]}{dt}\) (b) \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\)

Step 1: Analyzing Coefficients

The coefficients for this reaction are 4 for NH3, 5 for O2, 4 for NO, and 6 for H2O.

Step 2: Rates of Appearance and Disappearance

Using the coefficients, the rate expression can be written as: Rate = -\(\frac{1}{4}\) \(\frac{d[NH_3]}{dt}\) = -\(\frac{1}{5}\) \(\frac{d[O_2]}{dt}\) = \(\frac{1}{4}\) \(\frac{d[NO]}{dt}\) = \(\frac{1}{6}\) \(\frac{d[H_2O]}{dt}\) (c) \(2 \mathrm{C}_{2} \mathrm{H}_{2}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\)

Step 1: Analyzing Coefficients

The coefficients for this reaction are 2 for C2H2, 5 for O2, 4 for CO2, and 2 for H2O.

Step 2: Rates of Appearance and Disappearance

Using the coefficients, the rate expression can be written as: Rate = -\(\frac{1}{2}\) \(\frac{d[C_2H_2]}{dt}\) = -\(\frac{1}{5}\) \(\frac{d[O_2]}{dt}\) = \(\frac{1}{4}\) \(\frac{d[CO_2]}{dt}\) = \(\frac{1}{2}\) \(\frac{d[H_2O]}{dt}\) (d) \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{2}(g) \longrightarrow \mathrm{C}_{3} \mathrm{H}_{6}(g)+\mathrm{NH}_{3}(g)\)

Step 1: Analyzing Coefficients

In this reaction, all of the substances have a coefficient of 1.

Step 2: Rates of Appearance and Disappearance

Using the coefficients, the rate expression can be written as: Rate = -\(\frac{1}{1}\) \(\frac{d[C_3H_7NH_2]}{dt}\) = \(\frac{1}{1}\) \(\frac{d[C_3H_6]}{dt}\) = \(\frac{1}{1}\) \(\frac{d[NH_3]}{dt}\)

Key Concepts

Rate Expressions

In the study of reaction kinetics, a rate expression is a mathematical equation that shows the relationship between the rate of a chemical reaction and the concentrations of the reactants and products. It helps us understand how quickly or slowly a reaction proceeds over time. Rate expressions use the concept of rates of appearance and disappearance of substances in a chemical reaction.
For a general gas-phase reaction formula

• A + B \rightarrow C + D

the rate expression can be shown as:

• Rate = -\(\frac{1}{a}\) \(\frac{d[A]}{dt}\) = -\(\frac{1}{b}\) \(\frac{d[B]}{dt}\) = \(\frac{1}{c}\) \(\frac{d[C]}{dt}\) = \(\frac{1}{d}\) \(\frac{d[D]}{dt}\)

Here, the negative signs for reactants indicate their consumption over time, while positive signs for products show their formation. The terms \(\frac{d[A]}{dt}\), \(\frac{d[B]}{dt}\), \(\frac{d[C]}{dt}\), and \(\frac{d[D]}{dt}\) represent the change in concentration per unit time for each species. Dividing by their respective coefficients ensures that the rate calculation is consistent throughout the reaction.

Gas-Phase Reactions

Gas-phase reactions involve substances in the gaseous state interacting with one another to form new products. These types of reactions are important in fields like atmospheric chemistry and industrial processes. Understanding gas-phase reactions can help in developing clean energy technologies and reducing environmental pollution.
Characteristics of gas-phase reactions include:

• High molecular mobility: Particles are free to move and collide, increasing reaction rates.
• Volume and pressure dependency: Reaction rates can be affected by changes in pressure and volume.
• Homogeneity: All reactants and products are in the same state, making reactions comparably predictable.

In these reactions, the rate expression provides insight into how efficiently reactants transform into products under certain conditions. Observing the behavior of gases can involve measuring pressure changes or detecting concentration shifts.

Coefficients Analysis

Analyzing the coefficients in a chemical reaction is crucial when writing rate expressions. Coefficients indicate the number of moles of each substance involved. They help balance the chemical equation and ensure the mass is conserved across the reaction.
In rate expressions, coefficients play a role in adjusting rate calculations to reflect the stoichiometry of the reaction. For example, in the reaction

• \(2 \text{A} + 3 \text{B} \rightarrow \text{C} + 4 \text{D}\)

the coefficients \(2, 3, 1,\) and \(4\) tell us how the molecules relate to each other.
When writing rate expressions:

• Dividing the rate of change of concentration by the coefficient ensures the reaction rates of reactants and products are directly comparable, even if the number of molecules differs.
• Each term in the rate expression provides insight into how the change of one component might affect the entire reaction process, proportional to its coefficient.

By analyzing these coefficients, one can determine how the manipulation of reactant concentrations or reaction conditions might accelerate or decelerate the chemical process.